Problem: $f(n) = -4n$ $h(t) = 4t^{3}-4t^{2}-7t+7+3(f(t))$ $g(x) = 6x^{2}+x-3+4(h(x))$ $ g(h(-2)) = {?} $
First, let's solve for the value of the inner function, $h(-2)$ . Then we'll know what to plug into the outer function. $h(-2) = 4(-2)^{3}-4(-2)^{2}+(-7)(-2)+7+3(f(-2))$ To solve for the value of $h$ , we need to solve for the value of $f(-2)$ $f(-2) = (-4)(-2)$ $f(-2) = 8$ That means $h(-2) = 4(-2)^{3}-4(-2)^{2}+(-7)(-2)+7+(3)(8)$ $h(-2) = -3$ Now we know that $h(-2) = -3$ . Let's solve for $g(h(-2))$ , which is $g(-3)$ $g(-3) = 6(-3)^{2}-3-3+4(h(-3))$ To solve for the value of $g$ , we need to solve for the value of $h(-3)$ $h(-3) = 4(-3)^{3}-4(-3)^{2}+(-7)(-3)+7+3(f(-3))$ To solve for the value of $h$ , we need to solve for the value of $f(-3)$ $f(-3) = (-4)(-3)$ $f(-3) = 12$ That means $h(-3) = 4(-3)^{3}-4(-3)^{2}+(-7)(-3)+7+(3)(12)$ $h(-3) = -80$ That means $g(-3) = 6(-3)^{2}-3-3+(4)(-80)$ $g(-3) = -272$